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Differential equations - mixing problem

  1. Feb 11, 2007 #1
    A room containing 1000 cubic feet of air is originally free of carbon monoxide. Beginning at time t=0 cigarette smoke containing 4 percent carbon monoxide is blown into the room at 0.1 ft^3/min, and the well-circulated mixture leaves the room at hte same rate. Find the time when the concentration of carbon monoxide in the room reaches 0.012 percent.



    rate = rate in - rate out ?



    the 4% carbon monoxide part is really throwing me off. i don't exactly know what to do with this number. but other than that, i should be alright.

    CO enters : 0.04 x 0.1 OR 0.1, i'm not sure whether to factor in the 4% CO
    CO leaves : 0.04 x 0.1 S(t)/1000 OR 0.1 s(t)/1000

    i need to figure that part out.. after that, everything i can handle. can someone please help me with the 4% carbon monoxide part
     
  2. jcsd
  3. Feb 11, 2007 #2
    okay so i tried working it out with the 0.04 factored in as shown above... and at the end i get a ln -#, so i guess that is wrong.
    so now i am doing it without the 0.04. if i do it without the 0.04, then i do not understand the concentration part. it says to find the time when conc is 0.012 %, so set c(t)=0.00012 OR shoudl i be factoring in 4% CO somewhere still. i am really confused
     
  4. Feb 11, 2007 #3

    Pyrrhus

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    Here's a hint: make M(t) be the quantity of carbon monoxide in the
    room at any time and then the concentration is given by C(t)
    = M(t)/1000.
     
  5. Feb 11, 2007 #4
    i have used, s(t) = amount of CO
    c(t)=s(t)/1000

    so i have rate in as: 0.04 x 0.1
    rate out: 0.1 s(t)/1000

    i have modified my rate in and rate out from when i first posted the eq'n as i don't believe i have to add the 4% CO factor twice. tell me what you think now.
    i have solved this ALL the way to plugging in C(t)

    i plugged in c(t) as 1.2 x 10^-4
    my eq'n is
    c(t) = 4x 10^-10 (1-e^(1.0x10^-4)t)
    so pluggin in c(t)
    1.2 x 10^-4 = 4x 10^-10 (1-e^(1.0x10^-4)t)
    solving for t
    300000 = 1-e^(1.0x10^-4)t
    299999 = -e^(1.0x10^-4)t

    so.. once again. i am stuck.
     
  6. Feb 11, 2007 #5
    here is all my work.

    rate in: 0.04 x 0.01
    rate out: 0.1 x s(t)/1000

    s(0) = 0

    s'(t) = 0.004 - 1.0x10^-4 s(t)
    s'(t) + 1.0x10^-4 s(t) = 0.004

    solving diff eq'n:
    a(t) = 1.0x10^-4, b(t)=0.004

    using formula: u(t) = exp(integ(a(t)dt))
    u(t) = exp(integ(1.0x10^-4 dt))
    u(t)=exp(1.0x10^-4 t)

    using formula: d/dt (u(t) s(t) ) = u(t)b(t)
    d/dt ( (e^(1.0x10^-4 t )) s(t) ) = (e^(1.0x10^-4 t)) x 0.004
    (e^(1.0x10^-4 t )) s(t) = integ (0.004(e^(1.0x10^-4 t))dt)
    (e^(1.0x10^-4 t )) s(t) = 0.004(1.0x10^-4)(e^(1.0x10^-4 t) + C
    s(t) = [(4x10^-7) (e^(1.0x10^-4 t)) + C]/(e^(1.0x10^-4 t))]
    s(t) = 4x10^-7 + Ce^(1.0x10^-4 t)
    sub s(0)=0
    0 = 4x10^-7 + C
    C= -4x10^-7

    so,
    s(t)=(4x10^-7)(1-e^(1x10^-4 t))

    c(t) = s(t)/1000
    c(t) = (4x10^-10)(1-e^(1x10^-4 t))

    find t when c(t) = 1.2 x 10^-4
    1.2x10^-4 = (4x10^-10)(1-e^(1x10^-4 t))
    300000 = 1-e^(1x10^-4 t)
    299999 = -e^(1x10^-4 t)

    stuck. any suggestions or any wrong steps???
     
  7. Feb 12, 2007 #6

    Pyrrhus

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    Use natural logarithm.
     
  8. Feb 12, 2007 #7
    i end up ln-ing a negative..
     
  9. Feb 12, 2007 #8
    can anyone find the mistake?
    i have some help from the instructor as he said the percentage is percentage of the volume, so i guess that is the part that is wrong??? i'm not sure what he means by that.
     
  10. Feb 12, 2007 #9

    Pyrrhus

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    Re-check your steps. I already told you the idea

    [tex] \frac{dm(t)}{dt} = 0.1 * 0.04 - 0.1 \frac{m(t)}{1000} [/tex]

    Maybe you will have less mistakes in your calculations if you solved it like this.

    [tex] c(t) = \frac{m(t)}{1000} [/tex]

    [tex] 1000\frac{dc(t)}{dt} = 0.1 * 0.04 - 0.1 \frac{1000c(t)}{1000} [/tex]
     
  11. Feb 12, 2007 #10

    Pyrrhus

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    The mistake lies in these steps. You forgot the negative.
     
    Last edited: Feb 12, 2007
  12. Feb 12, 2007 #11
    i'm working it out your way now, but the negative i 4got to type in, but i had it on paper and it doesnt really make a difference as when i am plugging in s(0)=0, c is still the same regardless of that.
     
  13. Feb 12, 2007 #12

    Pyrrhus

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    It DOES make a difference for the logarithm.
     
  14. Feb 13, 2007 #13
    okay
    so here is my work:
    starting from pluggin in the concentration:
    s(t) = (4x10^-7) - (4x10^-7)e^(-1.0x10^-4 t)
    C(t) = s(t)/1000
    c(t) = (4x10^-7)(1-e^(-1.0x10^-4 t)) / 1000
    pluggin in c(t) = 1.2x10^-4
    1.2x10^-4 = 4x10^-10(1-e^(-1.0x10^-4 t))
    300000 = 1-e^(-1.0x10^-4 t)
    299999 = -e^(-1.0x10^-4 t)
    ln 299999 = ln -e^(-1.0x10^-4 t)
    i am stuck agian..
    at the same place
     
  15. Feb 13, 2007 #14

    Pyrrhus

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    Well i decided to do the problem to see why you weren't getting it right, anyway, i see another mistake, your integration is wrong. Btw, i get 30.05 as the answer, is t in minutes?, looks awfully fast if it was in seconds.

    [tex] \int e^{kx} dx = \frac{1}{k} e^{kx} + C [/tex]
     
    Last edited: Feb 13, 2007
  16. Feb 13, 2007 #15
    oh my god , thank you so much.
    I have checked over my work so many timse and still did not catch that!
    reworking my solution as we speak and hopefully i get the same answer as you.
     
  17. Feb 13, 2007 #16
    i got the same answer as you, thanks very much.
     
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